 Hello all welcome to the problem-solving session on important topics this particular session is basically meant for the J main and some J advance concepts So those who have joined in the chat Those who have joined in the session, please type in your names in the chat box Hello all good morning So without much delay, let us get started with the first problem of the day So here we have a question if one lies between the roots of this equation and Box x represents the greatest integer less than equal to x then the greatest integer function of 4 mod x by mod x square plus 16 whole to the power of m is equal to Is equal to which of the following options? So all good morning to you all please feel free to type in your response in the chat box Hello, Dipanjini Yes, guys any response? No idea All right So guys first of all if you look at this quadratic expression this quality expression has got the coefficient of y square as 1 That means it is a graph of a parabola which opens upwards Okay, it opens upwards Okay, and one lies between the roots. So if one lies between the roots Let's say x y equal to one lies between the roots It means the value of this function. Let's say I call this as y. Okay, so let me call this as y So f y is actually your term y square minus m y plus 1 Okay, so f 1 should be less than 0 if one lies Among between the roots alpha and beta If f 1 is less than 0 that means 1 minus m plus 1 should be less than 0 that means to Yes, now, can you see me? Can you see the screen now? Yeah, of course, there was a power cut for a small time is the screen visible am I audible to all of you now M is greater than 2 That means what this term over here is a value which is greater than 2 Let's see. How does it help us? Now all of these quantities that you see that is x mod x square 16 etc. These are all positive quantities. These are all positive quantities. So can I say arithmetic mean? greater than geometric mean if I apply I can say x square mod x square plus 16 by 2 would be greater than equal to 16 mod x square under root right Which means mod x square plus 16 would be greater than 4 mod x which implies 4 mod x by x square mod x square plus 16 will be less than equal to half Will be less than equal to half, right? Now if this is less than equal to half We are also aware that this will this will be a positive term as well. That means it will lie somewhere between 0 and half correct and If I raise it by a power which is greater than 2 If I raise both sides with a power which is greater than 2 Okay, then can I say this term this term would be still Lying somewhere between 0 and half because any fraction if you raise to a power which is greater than 1 It will still become smaller That means gif of this Gif of this expression would always be equal to 0 because it's a value which is somewhere between 0 and half Is that clear So of course Shruti was correct. She was the first one to answer this Option a will be correct in this case again, this has to strike to you this has to strike to you and How does it strike to you the presence of 16 and 4? mod x square and mod x Is that clear any question regarding this? Alright, so moving on to the second question Second question says the pair of state lines joining origin to the point of intersection of these two circles Yes, guys Any progress made on this? So you would have all learned the process of homogenization So homogenization is the process where let's say you have been given any curve and there's any line and You want to find out the pair of straight lines connecting this with the origin Okay, so when you want to find the pair of these two straight lines connecting the point of intersection of two curves by a line Okay, so here the question is find the pair of straight line joining origin to the points of intersection of these two circles So can I say the point of interest the line of intersection? The the line connecting the points of intersection of these two circles would be its common chord correct, so What are the equation of a common chord of? Circle whose equation is s dash equal to 0 and s double dash equal to 0 You can I say common chord will be nothing but s dash minus s double dash equal to 0 correct So when you subtract these two equations you will end up getting two gx plus fi To gx plus fi equal to minus of a square is Equal to minus of a square correct So let's say this is our line to gx plus Fy is equal to minus of a square right, and let us homogenize this with the Equation of the circle x square plus y square is equal to a square Okay So what I'll do next is I'll write a square into One square Okay, and instead of one square. I will write two gx plus two fy by Minus a square guys. Can you see the screen? Can you see the screen now? This which is four times gx plus gx plus fy Whole square by a to the power four Okay So when I replace it back over here what I will get is x square plus y square is equal to a square times four gx plus fy square By a to the power four So a square and it this will get cancelled off and you can simplify it as a square Can you see now? Okay, so finally this is what we get which is actually a square X square plus y square minus four gx plus fy whole square equal to zero Which is actually your option number B, which is correct. Sorry because of the constant power cut and all You may be facing some problem right now, but I think it's really fine in some time Right. So guys, please remember the concept of homogenization which actually helps you to find the Equation of the pair of lines connecting the intersection of a line and a curve of second-degree equation Homogenization doesn't work if the curve is having a General equation higher than degree two. So let's move on to the Third question today This question has been picked up from a 3d geometry the equation of a plane through the intersection of these two planes and parallel to this line So this is the Unsymmetrical or you can call it as asymmetrical form of the equation of a line. So when two planes are given basically the The line formed by the intersection of these two planes is written as the equation of two separate planes Which is called the asymmetrical form option C What about others? All right, so I think it's the easy question So a plane passing through the Intersection of these two planes and parallel to this line Please guys note that when y is zero and x is zero. It's saying that it's your x-axis right So So the equation of the desired plane will be this Now the equation is the equation of this plane is giving you a family member that we are looking for which is parallel to the x-axis That means the direction cosines of this plane directed direction cosines of the Norman to this plane will be Perpendicular to the direction cosines of x-axis So it's something like this. This is your plane and this is your x-axis Okay, so the Norman from the plane will be perpendicular to the x-axis so First of all grouping the terms containing Now we have one into a plus lambda a dash Plus zero into b plus lambda b dash Plus zero into c plus lambda c dash should be equal to zero Which means a plus lambda a dash is equal to zero, which means lambda comes out to be minus a by a dash Right, is that fine guys? Is this clear? So putting these values over here, why is it giving me zero everywhere and parallel to this line? Okay Lambda is correct guys, right? So when you put it back over here, I'll get the equation as ax plus by plus cz plus d Minus a by a dash a dash X b dash y Yes, yes, yes, that's what I'm wondering why there's no extra Yeah So you would realize that when you multiply with a dash throughout you will be getting something like this So why term will a x-dash term will totally vanish. You'll have a dash b minus a b dash Y plus z term will become a dash c minus a c dash Plus a dash d minus a d dash equal to zero So I think which option is matching with this Option number c is matching with this option number c is correct. So guys, let's move on to the fourth question Shweta, it is not telling you it is greater than that. It is telling you it is greater than equal to that Please read the question carefully. All right, so let's check here again as You rightly said this term Should always be greater than equal to this right So W plus x plus y plus z will always be greater than four times W x y z to the power of one by four Today's whole side to the power of four So it'll become two fifty six W x y z Now here it is the opposite way round. So here it is given. This is greater than equal to this. So This will only be possible if the equality holds true Right and aim will be equal to GM only when the quantities are all equal That means W x y z will all be equal to each other In other words if I use this equation If I use this equation, I will get 815 24 Let's say x or y or w whatever you want to take. This is equal to 24 So each one of them is going to be one. So w equal to x plus equal to y equal to z will all be equal to one Correct. Now using this you have to construct this new equation which of course will be one plus one which is two t square x into y will be again one t and And under root of four, which is two equal to zero Okay, and I have to find the nature of the roots of this equation So the nature of the roots of the equation will have to find the discriminant which is b square minus four ac Which is one minus four into two into two which is definitely negative Which means? Which means this particular equation has imaginary roots So option C is correct over here. Is that fine guys? So let us move on to the next question, which is the fixed question Alright, so Shweta said one Kushal says four Anybody else? Kushal, can you send me the snapshot of your working on my personal chat? So we have a parabola over here. Let's say I draw this parabola Y square is equal to 4x and a chord is drawn from a point P 1 comma t Okay, so which cuts the parabola at points a and b and it's given to us that P a into bb is 3 mod t Then we have to find the maximum value of it So what I will do is I will use the distance form of the line. So let's say This is at a distance of r from a P So I can write it as x minus 1 by cos theta Is equal to y minus t by sine theta is equal to r Okay So I can write clearly x as r cos theta plus 1 and y as r sine theta plus t Now since x and y that is this point x and y would lie on the parabola This would satisfy the equation of the parabola Y square is equal to 4x right So Y square Equal to 4x Y square is equal to 4x So let me simplify this it gives you r square sine square theta plus 2rt sine theta plus t square is Equal to 4r cos theta plus 4 which is going to be r square sine square theta plus r times 2t sine theta Minus 4 cos theta Plus t square minus 4 equal to 0 now we know for sure that there are two values of r possible So one will be let's say r1 and the other one will be let's say r2 So r1 and r2 would be the roots of this equation. So the roots of this equation would be r1 and r2 So the product of the two roots would be nothing but t square by 4 minus sine square theta And we'll take a mod over here. It just in case, you know, this becomes negative And this is equal to 3 mod t This is equal to 3 mod t which means This expression is equal to mod of sine square theta and we all know mod of sine square theta will be less than zero Sorry, it will be less than one So this expression will always be less than equal to okay Which means mod of t square minus 4 will always be less than equal to 3 mod t so we can write this as This and it is factorizable as so t lies between minus 1 and 4 T lies between minus 1 and 4 So if t lies between minus 1 and 4, the maximum value of t will be 4 So the maximum value of t will be 4. So yes, Kushal was correct Any question with respect to this guys? Please let me know. Is it clear? Please type CLR If it is clear. So let's move on to the next question That's question number 6 Sure, I'll go over the solution once again See first of all, what did I do is I use the distance form of the equation of a line? So let's say a line passing through P has the equation x minus 1 by cos theta y minus t by sine theta equal to r This is called the distance form of a line. This is called the distance form of a line. Okay So I've got my x and y in terms of theta r and t and then I use the fact that x and y would be these meeting points, right? So that would satisfy this parabola So when I got When I substituted x and y in the equation of a parabola, I got a quadratic in R and This quadratic in R has roots r1 and r2 Correct. What is given to us is pA into pB. pA into bB is actually r1 into r2 That is the product of the roots. Product of the root is C by A. This is your A This is your C So C by A. C by A will be this t square by 4 by sine square theta. That will be equal to 3 mod t Then I found sine square theta a mod is equal to this and we all know sine square theta The greatest value is 1. So this will be always be less than 1 I got an inequality. I solved the inequality and I got the maximum value of t as 4 Is it clear to you Shruti now? It's okay. No problem. Second option. Okay. What about others? Okay, option a Okay, anybody else I need at least okay psi s and d All right, so let's discuss this First of all, we need to find out the integral 0 to pi by 2 So for that you divide by cos square both the sides So we'll have secant square x dx and we'll have a square a b square tan square x And now take your tan x st So it becomes 0 to infinity dT by a square plus b square t square Taking your tan x st So secant square x dx will become dt So this will become 1 by b square 0 to infinity dt by t square a square by b square So this will be 1 by a 1 by a tan inverse of t by a okay It's going to be 1 by a b and this is going to be when you put the limits This is going to be pi by 2. So that's pi by 2 a b So pi by 2 a b is given to you as pi by 16, which means a b is equal to 8 Which means a b is equal to 8 Now, we know the minimum value of this is minus under root a square plus b square. Okay So a square plus b square by 2 will always be greater than under root of a b That means a square plus b square by 2 will always be greater than under root of 8 Sorry under root of a square b square. Yeah, sorry. So that will become Just root 8. Okay. So I can say a square plus b square will always be greater than 16 So under root of this will always be greater than this So a square plus b square under root will always be greater than 4. So the minimum value of this will be Minus under root a square plus b square, which is minus of 4. So which is going to be this So your option number a becomes correct So only Shweta gave the right answer over here. So guys read the question very very carefully I think this was an easy question. However, I should not have wasted time doing all these things as a serious Esperance, you are supposed to remember the result of this 5 by 2 a b Because this is so common integral that comes in your school exam as well. Is that fine any questions? All right, so I'll move on to the next question then question number seven any idea guys It's given that f of x is a quadratic expression x square minus a x plus b a is an odd positive integer and the roots of this equation are two distinct prime numbers If a plus b is 35 find the value of this expression Yeah, it's an integer. The answer will be an integer, but it's not an integer type All right, so let's discuss this A is an odd positive integer and the roots of this are two distinct prime numbers Okay, so let's say the two prime numbers are p and q So some of the root should be equal to a Correct and since these two are prime numbers These two are prime numbers and a is coming out to be odd Now prime numbers are normally known to be odd Right, but two odd numbers will never add to give you an odd number Correct, but two odd numbers will never add to give you an odd number Correct that means one of the prime numbers have to be even and the only even prime number is two correct The only even prime number is right So if one of the roots is two I Can always put this in two so f2 would be equal to zero which implies four minus two a plus b should be equal to zero That means two a minus b should be equal to four and a plus b is equal to 35 So three a is 39. So a is 13 So if a is 13 b has to be 22 correct, that means the quadratic expression f of x is x square minus 13 x Plus 22 equal to zero not equal to zero equal. This is equal to this Okay, so you can also factorize this as x minus 2 into x x plus 2 into x minus No, I don't think so it will be factorizable. Yeah minus 2 and minus 11 So this will be x minus 2 and x minus 11 This is your f of x correct Now what is f1? So let's figure it out f1 f1 would be 1 minus 2 by 1 minus 11 which is going to be minus 1 into minus 10, which is 10 F2 would be zero itself Similarly f3 would come out to be or you can treat this as a summation process You can treat this as a summation process because we cannot sit and literally count them up so I Can treat this as if you are trying to sum up You're trying to sum up x square minus 13 x plus 22 summation from x equal to 1 till 10 which is going to be N into n plus 1 into 2 n plus 1 which is by 6 minus 13 into Summation of this will be 10 into 11 by 2 plus 22 into 10 Okay, this multiplied with f of 10. So let me calculate this first. So this will be 2 7 This 5 so is 35 into 5 35 385 minus 55 into 11 in fact, I can simplify this here itself This will be 55 into 7. This will be 55 into 13 and this will be 55 into 4 so that will give you 55 into 7 plus 4 minus 13 which is going to be minus of 55 into minus of 2 which is minus 110 and what will be f of 10 f of 10 will be 10 minus 2 by 10 minus 1 11 which is going to be again minus of 1 minus of 8 right so my result would be Minus 110 into minus 8 which is plus 880 Which is plus 880. So absolutely correct. So those who answer 880. I think the first one to answer it was Maishnavi So moving on So this is a question which has been picked up from properties of triangles If O is the circum circle circum center of the triangle ABC and R1 R2 R3 are the radii of the circum circles of the triangle O, B, C, O, C, A and O, A, B respectively Then we have to find this expression Yes, any response from anyone? Not getting Okay So you guys first let us understand. Let's say this is a circle and we have these sites as a BNC okay The question is R1 is the radius of the circum circle of O, B, C So O, B, C, what is that? What is the formula for circum circles? Circum circle formula is ABC by 4 delta right So now let us under let us consider that let the area of Triangle O, B, C, B, delta 1 Okay So can I say R1 will be equal to ABC ABC for this will be R square a by 4 delta 1 Okay, that means a by R1 a by R1 will be 4 delta 1 by R2 Correct Similarly, similarly I can say B by R2 will be what? 4 delta 2 by R square and C by R3 will be 4 delta 3 by R square Okay Now when you add them that is a by R1, B by R2, C by R3 You are adding 4 R square delta 1 plus delta 2 plus delta 3 and The sum of all these delta is actually delta so it becomes 4 delta by R square right and again Delta is what? Delta is ABC by 4R So place it over here. So 4 times ABC by 4R into R square 4 4 gets cancelled You get ABC by R cube which is option number B is correct Option number B is correct guys properties of triangle you need to visit again I think the formulas have slipped out of your mind Okay, however, they're very easy. This is very easy concept. So now we can take a small break over here Let's have a small break and On the other side of the break We'll resume with more problems and we'll resume at 1145 a.m. All right, so let's resume back. Yeah, so let's take up this problem the ninth problem. Yes, guys Any response? Yeah, any response guys, please feel free to type it in the chat box This is a basic question on limits that you can expect See I use your common sense definitely if X is sending to 1 by alpha, you know that your denominator is going to be 0 right and Looking at your options. Your answer is not infinity means your numerator should also be 0 that means somehow This cubic equation should have one of its roots as 1 by alpha correct now clearly the product clearly you can see that if you put X as minus 1 it's going to be Satisfying it right it's going to be satisfying it. So other route is going to be minus 1 as well right and The product of the root is minus b by a c by minus d by a so let's say the other root is unknown to me Let's say beta. So this Into minus 1 into beta is also minus 1 that means beta is also alpha beta is also alpha So we have three roots now in front of us minus 1 alpha 1 by alpha Can you proceed from here on? Can you take it from here since we have known the roots for this now? So the roots are known for this Can you proceed from here on all right? So alpha is already given to us as the root Okay, so don't need to waste time during in doing this So this was not necessary So assume like this question this part of the question was not necessary So this let's say even if it is not given to you in the question You should be still be able to solve it. So you can write this as tan of A times x plus 1 x minus alpha x minus 1 by alpha Okay Now if you want to use your standard limit standard trigonometric limit if you want to use your standard trigonometric limit We have to multiply and divide with this term this gets cancelled and this becomes one correct, so you will be left with limit x Tending to 1 by alpha a x plus 1 x minus alpha By the way, this was alpha not a yeah by alpha okay, and Now when you substitute your values of x as 1 by alpha it becomes a 1 plus 1 by alpha 1 by alpha minus alpha by alpha So that becomes a Alpha plus 1 1 minus alpha square by alpha cube Which is going to be a alpha plus 1 square 1 minus alpha by alpha cube So option number D. Yes option number D is correct So option number D becomes correct overlay. Okay guys. I was expecting that he would solve these kind of problems And these are basic concepts which come out from limits, okay Moving on to the next question question number 10 for the day Just a hint you can use integration by parts over here. This is easy to integrate you can call this as V You can call this as you Minus one. Okay. No minus one is not the answer. Yeah That's not real. Yeah, that's not real for x greater than pi by 4. All right, so let's try this problem out I'll apply integration by parts over here Treating this as my u And treating this as my v. Okay So when I apply integration by parts, it's a u times integration of cosecant square cosecant square integration is minus of cot x ln Cos x plus under root of cos 2x okay minus Integral of a cosecant square is minus cot x times derivative of this term will be 1 by Cos x plus root of cos 2x Times the derivative of this term which is going to be minus of sine x plus Half cos 2x to the power minus half Into minus sine 2x Into 2 so it becomes Minus cot x numerator part. I'll just write it as sine x Plus 1 by 2 root cos x cos 2x into sine 2x Right the rest of the parts. I'll copy it as such now. Let us focus on simplifying this part So this will become integration of cot x I will have sine x a root of cos 2x Plus sine 2x by Cos x under root of cos 2x Whole into under root of cos 2x no doubt so far. So This time I'm copying as such So next what I'll do I'll write this as cos x by sine x. So that becomes integral of Cos x under root of cos 2x and this is sine 2 sine x Cos x so when it multiplies to Cos x by sine x it becomes 2 cos square x So here I'll get Minus 2 cos square x How can you further simplify this any idea? This is actually a huge problem. It would easily take around 10 to 15 minutes add and subtract one in the numerator, okay? How does that help? We can use integration by parts again in the third last step I think from here on we can simplify this we have to just simplify and try to try to bring out some terms Which are you know that can be integrable. So Okay, so let's see what happens when we add and subtract one. Okay, so Shweta is suggesting to add and subtract one to cos square x becomes cos 2x. Okay, so let's try doing this add and subtract one Okay, so let's try this out cos x under root of cos 2x and This becomes cos 2x Plus one then it becomes cos x under root cos 2x plus cos 2x so it becomes integral of one plus one by Cos x under root cos 2x plus cos 2x Yeah, now after that what to do No idea, right Okay, so after that we can just try out some you know conversion into a cos so Let's multiply cos x on top so it becomes cos square x and the root of cos 2x plus Cos x cos 2x Is this helping us out? Can we use x interval condition to evaluate the integral guys here? We have a this integral to solve So let's not bother about x right now. Let's treat it as an indefinite integrals to solve So we'll do one thing will try this out later on meanwhile. We'll move on with the other problem So we'll come back to this problem because I think some Integration step is something which we have to use over here So let us look into this question. Meanwhile, we'll come back to this So here we have a question where f of x is given as this cubic equation where a b c d are real and 3 b square is less than c squared This is an increasing function and g of x G of x is 8 times f dash x plus b times Double derivative of f of x plus c square and capital g of x is the integral of gt from alpha to x Now we have to comment about the nature of capital g of x More than one option may be correct in this So this question says f of x is this cubic equation and g of x is this so let us find g of x first So it's a times the derivative of this which is 3 a x square 2 b x plus c plus b times double derivative which is going to be a 6 a x plus 2 b and Plus c square so it's going to be 3 a square x square plus 2 a b and 6 a b which is 8 a b x and 2 b square Plus c square plus a c Okay Now let us first find the nature of g of x first of all g of x has its coefficient of x square as positive So this is positive Let's talk about the discriminant. Discriminant is b square b square minus 4 a C Okay Let us simplify this to a certain extent. I can take 4 a square common for sure So I'll have 16 b square 4 a square is gone. So minus 6 b square Minus 3 c square Minus 3 ac yeah So that will become 4 a square or 10 b square minus 3 c square minus 3 ac Okay Next we have also given that f of x has f of x is this expression and a b c d are real okay a b c d are real means The derivative of this function must have So let's find a derivative of this function f of x first which is going to be 3 a x square plus 2 b x plus c Okay, if this function is an increasing function means this term should always be positive Correct if this term is always positive means a should be always positive and the discriminant over here that is 4 b square minus 4 ac which is 12 ac should always be negative that means b square should always be less than 3 ac correct So b square should always be less than 3 ac. So can I write this expression to be less than? Instead of 10 b square Sorry instead of ac I can write this term as b square itself So I replace 3 ac term with b square because I wanted to come to this expression since b square is less than 3 ac Minus 3 ac will be greater than B square will be greater than minus 3 ac So what I did I I replace my b square with or replace my minus b square with or 3 ac with minus b square Okay Which means this expression for a square? This will become 9 b square minus 3 c square which is Take 3 out so 3 into 4 a square This will be 3 square minus c square So is it clear what I've done So after this step when I replace 3 ac with b square this this quantity becomes lesser than this because b square is lesser than 3 ac so you are instead of subtracting a higher quantity you are subtracting a lower quantity So the resultant expression should be higher than the previous one correct so that implies 12 a square 3 b square minus c square and we know that 3 b square is lesser than 3 b square is lesser than c square. So this is going to be a negative term This is going to be a negative term. This is going to be a positive term So this entire quantity is going to be less than zero that means Discriminate determinant of g dash x is less than zero means g of x is always positive Correct, so if g of x is always positive For a x all belonging to our what I can I commend about capital g of x which is actually integral of Which is actually integral of G of t dt from alpha to x So let me just use Leibniz rule over here differentiate it you get g of x and since g of x is always positive It implies g dash x will also be positive. That means g of x is an increasing function G of x is an increasing function So I think option number b is definitely correct But if option b is correct d will also be correct because increasing function. It has to be a one-one function as well Is that fine guys any question? This is a typical j advance level question. Is that fine alright? So moving on to the 12th question So next question in front of you is a vector question a is given to you b is given to you And x1 x2 x3 can take only values from minus 3 to 2 number of possible vectors b says that a and b are mutually perpendicular Okay, 1918 What about the others? Okay, so Most people are getting 19 and 18 only Alright, so let's discuss this So when you do a dot b You get x1 plus x2 plus x3 correct So when you take the dot product of these two vectors you get x1 plus x2 plus x3 and this should be zero Okay Now you have to choose Such values of x1 x2 x3 says that you get zero and you have only these many options Okay, so can I use multinomial theorem over here? Can I use multinomial theorem here? That means can I use the fact that I need to get the I need to get the Coefficient of x to the power zero in this That means I need to get the coefficient of 1 plus x plus x square plus x cube plus x4 plus x5 By x cube to the power of 3 I need to get the coefficient of x to the power zero in this Right in other words, I need to get the coefficient of x to the power nine in this Because in the denominator, I'll get x to the power nine right So if I'm able to find the coefficient of x to the power nine from this my job will be done Correct. Now. This is a gp This is a gp Whose first term is one common difference is x. So basically I'm looking for coefficient of x to the power nine in this That means coefficient of x to the power nine in this expression Now this can be only be written till 1 minus x to the power six because other terms will exceed x to the power nine So let me stop here And this will be infinitely, you know begin number. Okay, so this is like 1 minus x to the power 1 minus x to the power Minus n right which we know is expanded as 1 plus nx Plus n n plus 1 by 2 factorial x square and so on Okay, now I'll hand pick up terms from here. So first of all, I need the x to the power nine term from here So what are terms are required from here x to the power nine term? What is x to the power three term? Okay now What is going to be the coefficient of x to the power r in this coefficient of x to the power r in this is just going to be n n plus 1 on the way till n plus r minus 1 x to the power r by r factorial Which is actually n plus r minus 1 factorial by r Factorial n minus 1 factorial. That's actually n plus r minus 1 cr So remember n over here is 3 And r for this is 9 So coefficient of x to the power 9 would be what coefficient of x to the power 9 would be 9 plus 3 minus 1 which is 11 c 9 Okay So this will be 11 c 9 Coefficient of x to the power 3 will be Coefficient of x to the power 3 will be 3 plus 3 minus 1 c 3 which is going to be 5 c 3 So this is 5 c 3 So your final answer will be 1 into x to the power 9 which is 11 c 9 Okay Oh one mistake. I mean this should be 3 over here. I'm sorry. Yeah So the other coefficient will be minus 3 into x to the power 3 coefficient which is 5 c 3 So swetha is asking had it been x 1 plus 3 x 2 plus x 3 equal to 0 Then what will be the change in the multinomial theorem? See in that case, uh, what I will do for x 2. I will always I will always cube all those all those terms Are you getting it just for the middle expression x 2 I will be using the powers like see whatever is the value of x 2 that will always make it three times So it can take values like minus 9 minus 6 Minus 3 0 so that those terms I will be using it Why because x 2 can take values which are minus 3 minus 2 minus 2 minus 1 0 1 2 So 3 x 2 will take triple those values correct So for the middle term For the other terms I would write this For the other terms I would be writing this But for the middle term I would be writing x to the power minus 9 x to the power minus 6 Etc till x to the power. Sorry. This will be plus 2 got it swetha right So, chalo will complete this question over here By the way, this was not required So we'll complete the question over here. So 11 c 9 11 c 9. Let's calculate its 10 11 into 10 By 2 minus 3 into 10, which is this So that's going to be 55 minus 30 55 minus 30 means Answer will be 25 So none of you could get it right So moving on to the next problem This is a question from ellipse So the question is the equation of the family of ellipse is given to you then the locus of the extremities of the latter spectrum You have to find the locus of the extremities of the latter sector I wrote it on the device itself Any response guys for this So first of all your a square is equal to cos square alpha Correct and b square is sin square alpha. So we know that This is equal to 1 But we also know that a square is b square is a square 1 minus c square Now, why we know this because in the interval 0 to pi by 4, this will be at a higher level than this So a would be greater than b So the formula will be b square is a square 1 minus e square correct That means a square plus a square 1 minus e square is equal to 1 So 2 a square Minus 1 is a square e square Now, what is the extremities of the latter sector? What is the extremities of the latter sector? In an ellipse So if you draw an ellipse What would be the extremities of the latter sector? Can I say it's a a comma b square by And minus a sorry a comma minus b square by This will be minus a comma b square by a and minus a comma minus b square by Is that okay? Any question so far? So, uh, I can say safely that x could be Plus minus a e and y could be plus minus b square by Right. That means x could be plus minus under root 2 a square minus 1 and y will be Plus minus now b square is from here. I can say b square is 1 minus a square So this is 1 minus a square by Okay Now from these two I have to try to eliminate my a I have to try to eliminate my a Okay, so what I'll do is I'll write x square as 2 a square minus 1 so x square plus 1 is 2 a square Correct from here. I can say Uh Y square is 1 minus a square whole square by a square correct So replacing a square with x square plus 1 by 2 in both the positions So replacing it over here. I get y square is equal to 1 minus a square whole square By a square So let's let's bring the a square on top. So it becomes x square plus 1 by 2 So that becomes x square plus 1 y square by 2 is equal to this becomes 1 minus x square whole square by 4 So this 2 and this 2 gets cancelled. So it becomes 2 x square plus 1 y square 1 minus x square whole square So your option number option number option number B becomes correct Option number B becomes correct. So guys, uh Only I think the integration question is left for us to be solved So I would request you all to try it again. Okay. Well, we may have done some silly mistakes or some errors over here but, uh, please check this and Treat this as a homework question Okay, and try to solve this Sure Shweta, I'll just repeat the last part as well see once I got the Once I got x equal to a plus minus a and y equal to plus minus a I try to get everything in terms of a So plus minus a from here is nothing but plus minus under root 2 a square minus 1 So this is your x This is your y Correct. So x square is this y square is this So here I made a is is something which I assumed so a is a parameter I don't want any parameter to occur in my final answer. So I replace a in terms of x and put it back in the other equation Okay, so a square is this and I put it in in this equation of y square And then I simplified it and I got my final answer. Is that clear Shweta? So request you all to complete that integration problem and please post the solution on the group if you have found it Okay Read that as homework Any exams you have on monday or exams are over Oh practicals are left. Okay. So all right guys Thank you so much for coming online Okay over and out from my side. Bye. Bye And best of luck for your practicals